# need help again! TRIG FNS AND GRAPHS (1 Viewer)

#### MinnieMinnie

##### Member

I know how to do everything but (c)
The answer is supposed to be 0.52

(a) 1.91
(b) 0.95m
(d) 0.5

How would i work out c?

#### idkkdi

##### Active Member
View attachment 28412
I know how to do everything but (c)
The answer is supposed to be 0.52

(a) 1.91
(b) 0.95m
(d) 0.5

How would i work out c?
sin-1(0.95/1.91)

tbh my intuition says its that value x2.

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#### CM_Tutor

##### Well-Known Member
At $\bg_white t = 0$, $\bg_white y = A \sin b = 0.95$ and you know that $\bg_white A = 1.91$, so solve:
$\bg_white \sin b = \frac{0.95}{1.91}$.

#### idkkdi

##### Active Member
At $\bg_white t = 0$, $\bg_white y = A \sin b = 0.95$ and you know that $\bg_white A = 1.91$, so solve:
$\bg_white \sin b = \frac{0.95}{1.91}$.
Hey where's my logic wrong here. One period of the graph is 720 degrees. So 0.52x2 = 1.04 for b?

#### CM_Tutor

##### Well-Known Member
Hey where's my logic wrong here. One period of the graph is 720 degrees. So 0.52x2 = 1.04 for b?
This needs to be done in radians. The value of b gives the shift of an x intercept away from the origin.

#### idkkdi

##### Active Member
This needs to be done in radians. The value of b gives the shift of an x intercept away from the origin.
ah ok, that makes sense. But doesn't it look like its shifted by 104 degrees since it's shifted by 2/3 of the first quarter of a period.

#### CM_Tutor

##### Well-Known Member
The zero point is at
$\bg_white nt + b = 0 \quad \implies \quad t = \frac{-b}{n}$
which corresponds to your idea, but it doesn't alter the value of $\bg_white b$ itself which was found when $\bg_white t = 0$ and so was not dependent on $\bg_white n$.

#### idkkdi

##### Active Member
The zero point is at
$\bg_white nt + b = 0 \quad \implies \quad t = \frac{-b}{n}$
which corresponds to your idea, but it doesn't alter the value of $\bg_white b$ itself which was found when $\bg_white t = 0$ and so was not dependent on $\bg_white n$.
y=sin(nt+b)
= sin(n(t+b/n))

At 0.5, b/n = 2b. Inside the brackets is t+2b, therefore the graph of t+2b changes by 2b meaning that if a graph where n =1 is sketched out, it changes by two b. Now it is a matter of applying the multiple n, which basically horizontally shrinks the graph by a factor of 2, thereby making the change only b. However, as it is horizontally shrunk, it appears as if the same proportion of the curve for one period of the curve has been moved.

#### CM_Tutor

##### Well-Known Member
y=sin(nt+b)
= sin(n(t+b/n))

At 0.5, b/n = 2b. Inside the brackets is t+2b, therefore the graph of t+2b changes by 2b meaning that if a graph where n =1 is sketched out, it changes by two b. Now it is a matter of applying the multiple n, which basically horizontally shrinks the graph by a factor of 2, thereby making the change only b. However, as it is horizontally shrunk, it appears as if the same proportion of the curve for one period of the curve has been moved.
If I take the question as having the function: $\bg_white y = A\sin(nt + b)$ with $\bg_white A = 1.91\text{ m}$ and given that at $\bg_white t = 0\text{ h}$ we have $\bg_white y = 0.95\text{ m}$ then:

\bg_white \begin{align*} 0.95 &= 1.91\sin(0 + b) \\ \frac{0.95}{1.91} &= \sin b \\ b &= \sin^{-1}\frac{0.95}{1.91} \\ &=0.52057... \end{align*}

The period of the curve is $\bg_white T = 12.37 \text{ h}$ and since

$\bg_white T = \frac{2\pi}{n} \implies n = \frac{2\pi}{T} = \frac{2\pi}{12.37} = 0.507937...$

Thus, we have $\bg_white y = 1.91\sin(0.507937...t + 0.52057...)$

At 3 am, $\bg_white t = 3 \text{ h}$ and the tide is at $\bg_white y = 1.91\sin(0.507937... \times 3 + 0.52057...) = 1.91\sin 2.044... \approx 1.70 \text{ m}$

And, the height was $\bg_white y = 0$ prior to midnight when:

\bg_white \begin{align*} 0 &= 1.91\sin(0.507937...t + 0.52057...) \\ 0 &= \sin(0.507937...t + 0.52057...) \\ 0 &= 0.507937...t + 0.52057... \\ t &= \frac{-0.52057...}{0.507937...} \approx -1\text{ h } 2 \text{ min} \end{align*}

So, the height was at $\bg_white y = 0$ at 1 h 2 min before midnight, or 10:58 pm.

---

Now, I can do the same process with a model $\bg_white y = A\sin n(t + b)$ and reach the same conclusions, only with a different value of $\bg_white b$.

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If I take the question as having the function: $\bg_white y = A\sin n(t + b)$ with $\bg_white A = 1.91\text{ m}$ and given that at $\bg_white t = 0\text{ h}$ we have $\bg_white y = 0.95\text{ m}$ then:

\bg_white \begin{align*} 0.95 &= 1.91\sin n(0 + b) \\ \frac{0.95}{1.91} &= \sin nb \\ nb &= \sin^{-1}\frac{0.95}{1.91} \\ &= 0.52057... \end{align*}

The period of the curve is $\bg_white T = 12.37 \text{ h}$ and since

$\bg_white T = \frac{2\pi}{n} \implies n = \frac{2\pi}{T} = \frac{2\pi}{12.37} = 0.507937...$

from which we can now find $\bg_white b$:

\bg_white \begin{align*} nb &= 0.52057... \\ 0.507937...b &= 0.52057... \\ b &= \frac{0.52057...}{0.507937...} = 1.024... \end{align*}

Thus, we have $\bg_white y = 1.91\sin 0.507937...(t + 1.024...)$

At 3 am, $\bg_white t = 3 \text{ h}$ and the tide is at $\bg_white y = 1.91\sin0.507937... (3 + 1.024...) = 1.91\sin 2.044... \approx 1.70 \text{ m}$

And, the height was $\bg_white y = 0$ before midnight when:

\bg_white \begin{align*} 0 &= 1.91\sin0.507937...(t + 1.024...) \\ 0 &= \sin0.507937...(t + 1.024...) \\ 0 &= 0.507937...(t + 1.024...) \\ 0 &= t + 1.024... \\ t &= -1.024... \approx -1\text{ h } 2 \text{ min} \end{align*}

So, the height was at $\bg_white y = 0$ at 1 h 2 min before midnight (to the nearest minute), or at 10:58 pm.

----

By either method:
• the $\bg_white t$-intercept before $\bg_white t = 0$ occurs at $\bg_white t = -1.024\text{ h}$
• the $\bg_white y$-intercept occurs at $\bg_white A = 1.91\text{ m}$
• the amplitude is $\bg_white A = 1.91\text{ m}$
• the period is 12.37 h
because the functions $\bg_white y = 1.91\sin 0.507937...(t + 1.024...)$ and $\bg_white y = 1.91\sin (0.507937...t + 0.52057...)$ are the same.

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#### CM_Tutor

##### Well-Known Member
View attachment 28412
I know how to do everything but (c)
The answer is supposed to be 0.52

(a) 1.91
(b) 0.95m
(d) 0.5

How would i work out c?
MinnieMinnie, may I ask where this came from?

Everyone should note that the way this curve is drawn, with apparent vertical inflexions at each $\bg_white t$-intercept, is not how a sine function should be drawn.
\bg_white \begin{align*} y &= \sin \theta \\ \frac{dy}{d\theta} &= \cos \theta \\ \text{If \sin \theta = 0 then \theta = n\pi and so } m_{tang} &= \frac{dy}{d\theta} = \cos n\pi = \pm 1 \end{align*}
making the tangents at the intecepts diagonal.

In this specific case:

\bg_white \begin{align*} y &= 1.91\sin (0.51t + 0.52) \\ \frac{dy}{dt} &= 1.91\cos (0.51t + 0.52) \times 0.51 \\ &= 0.97\cos(0.51t + 0.52) \\ \text{If \sin (0.51t + 0.52) = 0 then 0.51t + 0.52 = n\pi for n \in \mathbb{Z} and so } m_{tang} &= \frac{dy}{dt} = 0.97\cos n\pi = \pm 0.97 \end{align*}

and so the $\bg_white t$-intercepts are close to having a gradient of 1... which is not at all how the diagram has been drawn by the source.

#### CM_Tutor

##### Well-Known Member
To illustrate, attached is a more accurate picture of this function...

Note that the tangents at the $\bg_white t$-intercepts have gradients near $\bg_white \pm1$ as indicated by the above calculations and fitting with the shift of -1.024 from the origin.

#### MinnieMinnie

##### Member
MinnieMinnie, may I ask where this came from?

Everyone should note that the way this curve is drawn, with apparent vertical inflexions at each $\bg_white t$-intercept, is not how a sine function should be drawn.
\bg_white \begin{align*} y &= \sin \theta \\ \frac{dy}{d\theta} &= \cos \theta \\ \text{If \sin \theta = 0 then \theta = n\pi and so } m_{tang} &= \frac{dy}{d\theta} = \cos n\pi = \pm 1 \end{align*}
making the tangents at the intecepts diagonal.

In this specific case:

\bg_white \begin{align*} y &= 1.91\sin (0.51t + 0.52) \\ \frac{dy}{dt} &= 1.91\cos (0.51t + 0.52) \times 0.51 \\ &= 0.97\cos(0.51t + 0.52) \\ \text{If \sin (0.51t + 0.52) = 0 then 0.51t + 0.52 = n\pi for n \in \mathbb{Z} and so } m_{tang} &= \frac{dy}{dt} = 0.97\cos n\pi = \pm 0.97 \end{align*}

and so the $\bg_white t$-intercepts are close to having a gradient of 1... which is not at all how the diagram has been drawn by the source.
This was on our HW sheet!!!

#### CM_Tutor

##### Well-Known Member
This was on our HW sheet!!!
Thanks for replying, MinnieMinnie. Wherever your school / teacher got the question from, the illustration of the trig function is awful.